Let
- $(\Omega,\mathcal A,\operatorname P)$ be a complete probability space
- $(X_t)_{t\ge0}$ be a time-homogeneous Markov process on $(\Omega,\mathcal A,\operatorname P)$ and $\kappa_t$ denote a regular version of the conditional distribution of $X_t$ given $X_0$ for $t\ge0$, i.e. $$\kappa_t(x,B)=\operatorname P\left[X_t\in B\mid X_0=x\right]\;\;\;\text{for all }(x,B)\in\mathbb R\times\mathcal B(\mathbb R)\tag1$$
- $C_0(\mathbb R)$ denote the space of continuous functions on $\mathbb R$ vanishing at infinity
- $\mu$ be a probability measure on $(\mathbb R,\mathcal B(\mathbb R))$
As usual, let $$\kappa_tf:=\int\kappa_t(x,{\rm d}y)f(y)$$ for bounded Borel measurable $f:\mathbb R\to\mathbb R$ and $t\ge0$. Assume $$\kappa_tC_0(\mathbb R)\subseteq C_0(\mathbb R)\;\;\;\text{for all }t\ge0.\tag2$$
Let $$\mu_t(B):=\operatorname P\left[X_t\in B\right]\;\;\;\text{for }B\in\mathcal B(\mathbb R)$$ for $t\ge0$. It's easy to see that weak convergence of $\mu_t$ to $\mu$ as $t\to\infty$ implies that $\mu$ is invariant with respect to $(\kappa_t)_{t\ge0}$, i.e. $$\mu\kappa_t=\mu\;\;\;\text{for all }t\ge0,$$ where $\mu\kappa_t$ denotes the composition of $\mu$ and $\kappa_t$.
Does the converse hold as well, i.e. if $\mu$ is invariant with respect to $(\kappa_t)_{t\ge0}$, are we able to conclude that $\mu_t$ converges weakly to $\mu$ as $t\to\infty$?